On Tue, 14 Apr 2009, jimm-pa...@gmx.de wrote:
Hi all,
I'm fitting a line to my dataset. Later I want to predict missing values that exceed the
[min,max] interval of my empirical data, therefore I choose surface="direct"
for extrapolation.
l1<-loess(y1~x1,span=0.1,data.frame(x=x1,y=y1),contro
-
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Stavros Macrakis
Sent: Tuesday, April 14, 2009 2:15 PM
To: jimm-pa...@gmx.de
Cc: r-help@r-project.org
Subject: Re: [R] Forcing the extrapolation of loess through the origin
On
Hi Torsten,
If you are fitting a line, why are you using "loess"? Why not simply
use "lm" to fit a regression line that goes through the origin? (i.e.
with no intercept).
Julian
jimm-pa...@gmx.de wrote:
Hi all,
I'm fitting a line to my dataset. Later I want to predict missing values that
On Tue, Apr 14, 2009 at 1:08 PM, wrote:
> I'm fitting a line to my dataset. Later I want to predict missing values that
> exceed the [min,max] interval of my empirical data, therefore I choose
> surface="direct" for extrapolation.
>
> l1<-loess(y1~x1,span=0.1,data.frame(x=x1,y=y1),control=loess
extrapolation of loess through the origin
Hi all,
I'm fitting a line to my dataset. Later I want to predict missing values
that exceed the [min,max] interval of my empirical data, therefore I choose
surface="direct" for extrapolation.
l1<-loess(y1~x1,span=0.1,data.frame(x=x1,y=y1),con
Hi all,
I'm fitting a line to my dataset. Later I want to predict missing values that
exceed the [min,max] interval of my empirical data, therefore I choose
surface="direct" for extrapolation.
l1<-loess(y1~x1,span=0.1,data.frame(x=x1,y=y1),control=loess.control(surface="direct"))
In my applica
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